Solvable and Nilpotent Lie Algebras — Examples

Since $\mathfrak{a}$ is abelian, ${\gamma }_2(\mathfrak{a})=[\mathfrak{a},\mathfrak{a}]=0$. If $\mathfrak{a}\mathrm{\ne }0$, then ${\gamma }_1(\mathfrak{a})=\mathfrak{a}\mathrm{\ne }0$, so the least vanishing index is $2$ and [L1] gives class one.

If $\mathfrak{a}=0$, already ${\gamma }_1(\mathfrak{a})=0$, and the explicit convention in [L1] gives class zero rather than one. These are the only two cases, and the calculation uses no choice.

Bilinearity, alternation, and centrality of $z$ show that every bracket is a scalar multiple of $z$, and the bracket $[x,y]=z$ realizes every such multiple. Hence ${\gamma }_2({\mathfrak{h}}_3)=[{\mathfrak{h}}_3,{\mathfrak{h}}_3]=kz\mathrm{\ne }0$.

Since $z$ is central, ${\gamma }_3({\mathfrak{h}}_3)=[{\mathfrak{h}}_3,kz]=0$. Thus the lower central series is ${\mathfrak{h}}_3,kz,0$; its second term is nonzero and its third is zero, so [L1] gives nilpotency class exactly two. The computation works in every characteristic and uses no choice.

For $p\ge 1$, let $F_p$ be the span of $E_{ij}$ with $j-i\ge p$; thus $F_1={\mathfrak{n}}_n$ and $F_n=0$. The identity $E_{ij}{E}_{\mathrm{\ell }m}={\delta }_{j\mathrm{\ell }}{E}_{im}$ shows that $[F_p,F_q]\subseteq F_{p+q}$: either product is zero, or its surviving matrix unit has superdiagonal distance equal to the sum of the two input distances.

Induction using [L1] and step 1.1 gives ${\gamma }_r({\mathfrak{n}}_n)\subseteq F_r$. Conversely, $F_1={\gamma }_1$, and if $r\ge 2$ and $E_{ij}\in F_r$, then $j-i\ge r$ and $E_{ij}=[E_{i,i+1},E_{i+1,j}]$; induction puts $E_{i+1,j}\in F_{r-1}={\gamma }_{r-1}$, so $E_{ij}\in {\gamma }_r$. Hence ${\gamma }_r({\mathfrak{n}}_n)=F_r$ for every $r\ge 1$.

In particular, ${\gamma }_n=F_n=0$, while ${\gamma }_{n-1}=F_{n-1}=k{E}_{1n}\mathrm{\ne }0$. Equivalently, the explicit left-normed chain $[E_{12},E_{23},\dots ,E_{n-1,n}]=E_{1n}$ witnesses sharpness. Thus [L2] gives class $n-1$. When $n=2$, the chain has one entry and ${\mathfrak{n}}_2=k{E}_{12}$ is abelian of class one; for $n=1$, outside the stated range, ${\mathfrak{n}}_1=0$ has class zero. No choice is used.

The diagonal entry of a commutator of two upper triangular matrices is zero, so ${\mathfrak{b}}_n^{(1)}$ lies in the strictly upper triangular subspace $F_1$. More generally, if $F_p$ is spanned by $E_{ij}$ with $j-i\ge p$, matrix-unit multiplication gives $[F_p,F_q]\subseteq F_{p+q}$ and $F_n=0$. Induction yields ${\mathfrak{b}}_n^{(r)}\subseteq F_{2^{r-1}}$ for $r\ge 1$, so a derived term vanishes once $2^{r-1}\ge n$. Thus ${\mathfrak{b}}_n$ is solvable by [L1].

Let $H=E_{11}$ and $X=E_{12}$. Then $[H,X]=X$, so $X\in {\gamma }_2({\mathfrak{b}}_n)$. If $X\in {\gamma }_r$, the same bracket puts $X$ in $[{\mathfrak{b}}_n,{\gamma }_r]={\gamma }_{r+1}$. Hence $X$ belongs to every ${\gamma }_r$ for $r\ge 2$, and none of these terms is zero. Therefore ${\mathfrak{b}}_n$ is not nilpotent by [L2].

Steps 1.1 and 1.2 prove the two promised properties over every field. The restriction $n\ge 2$ is sharp for this conclusion: ${\mathfrak{b}}_1=k{E}_{11}$ is one-dimensional abelian and therefore nilpotent. All witnesses are explicit and no choice is used.

Every basis bracket is zero or a scalar multiple of $y$, and $[x,y]=y$ is nonzero. Hence $[\mathfrak{a},\mathfrak{a}]=ky$, while $[ky,ky]=0$. Therefore ${\mathfrak{a}}^{(1)}=ky$ and ${\mathfrak{a}}^{(2)}=0$, so $\mathfrak{a}$ is solvable by [L1].

The same first calculation gives ${\gamma }_2(\mathfrak{a})=ky$. Since $[\mathfrak{a},ky]=ky$, induction gives ${\gamma }_r(\mathfrak{a})=ky\mathrm{\ne }0$ for every $r\ge 2$. Thus the lower central series does not terminate and $\mathfrak{a}$ is not nilpotent by [L2].

These computations establish every displayed equality and both conclusions over every field, including characteristic two, because the persistent bracket has coefficient one. No choice is used.

The standard rotation action gives $[R,P]=Q$, $[R,Q]=-P$, and $[P,Q]=0$ via [L1]. Therefore every bracket lies in the translation ideal $T=\mathbb{R}P\oplus \mathbb{R}Q$, while the first two displayed brackets span $T$. Hence $\mathfrak{e}(2{)}^{(1)}=T$. Since $T$ is abelian, $\mathfrak{e}(2{)}^{(2)}=0$. The first derived term is nonzero, so [L2] gives derived length exactly two.

The same calculation gives ${\gamma }_2(\mathfrak{e}(2))=T$. Moreover $[\mathfrak{e}(2),T]=T$, because bracketing $R$ with $P,Q$ again yields $Q,-P$. Induction gives ${\gamma }_r(\mathfrak{e}(2))=T\mathrm{\ne }0$ for every $r\ge 2$, so the algebra is not nilpotent by [L3].

Thus the abelian translation ideal is responsible for solvability, while the nontrivial rotation action prevents lower-central termination. Both endpoint computations are exact, and the finite explicit example uses no choice.

Put $U=\mathrm{span}(e_2,\dots ,e_n)$. Then $[U,U]=0$ and $[e_1,U]\subseteq U$. For a Jacobi triple entirely in $U$ every term is zero. For a triple containing exactly one copy of $e_1$, each possibly nonzero inner bracket lies in $U$ and is then bracketed with an element of $U$, so every term is zero. If a triple contains at least two copies of $e_1$, the two possibly nonzero terms cancel by bilinearity and the alternating law. Thus Jacobi holds, including in characteristic two, and the displayed rule defines a Lie algebra.

Every nonzero basis bracket is one of $e_3,\dots ,e_n$, and all of these occur. Hence ${\mathfrak{f}}_n^{(1)}=\mathrm{span}(e_3,\dots ,e_n)$. This subspace lies in the abelian space $U$, so ${\mathfrak{f}}_n^{(2)}=0$. Since $e_3\mathrm{\ne }0$ for $n\ge 3$, the least vanishing derived index is two by [L1].

The first bracket span gives ${\gamma }_2=\mathrm{span}(e_3,\dots ,e_n)$. If $2\le r\le n-2$ and ${\gamma }_r=\mathrm{span}(e_{r+1},\dots ,e_n)$, then only bracketing with $e_1$ contributes, and the displayed rule gives ${\gamma }_{r+1}=\mathrm{span}(e_{r+2},\dots ,e_n)$. Induction proves the promised formula through ${\gamma }_{n-1}=k{e}_n\mathrm{\ne }0$; finally $[{\mathfrak{f}}_n,k{e}_n]=0$, so ${\gamma }_n=0$. By [L2] and [L3], the class is exactly $n-1$.

When $n=3$, the calculation reads ${\gamma }_2=k{e}_3\mathrm{\ne }0$ and ${\gamma }_3=0$, so the smallest permitted dimension has class two and derived length two. For every $n\ge 3$, steps 2.1 and 2.2 establish both exact endpoints using only the displayed finite basis; no choice principle is used.

By [L3], $\mathfrak{a}$ is solvable. It is an ideal of itself, so the universal property [L1] gives $\mathrm{rad}(\mathfrak{a})=\mathfrak{a}$.

The line $ky$ is an ideal because $[x,y]=y$ and $[y,y]=0$. It is abelian and therefore nilpotent, so [L2] gives $ky\subseteq \mathrm{nilrad}(\mathfrak{a})$.

Let $I$ be an ideal not contained in $ky$. It contains a vector $v=ax+by$ with $a\mathrm{\ne }0$. Since $I$ is an ideal, $[v,y]=ay$ lies in $I$, whence $y\in I$; then $x=a^{-1}(v-by)\in I$. Thus $I=\mathfrak{a}$, which is not nilpotent by [L3]. Consequently every nilpotent ideal is contained in $ky$.

Step 1.2 shows that $ky$ is a nilpotent ideal, and step 1.3 shows that it contains every nilpotent ideal. The universal property [L2] therefore gives $\mathrm{nilrad}(\mathfrak{a})=ky$. Together with step 1.1 this proves both displayed identities. The only division is by the explicitly nonzero scalar $a$; no choice principle is used.

The line $ky$ is an ideal because $[x,y]=y$ and $[y,y]=0$. The displayed inclusion is injective, the quotient map is surjective, and its kernel is exactly $ky$, so the sequence is short exact.

Nevertheless, [L1] says that every lower-central term of $\mathfrak{a}$ from ${\gamma }_2$ onward is the nonzero line $ky$. Thus the ambient algebra is not nilpotent.

The kernel $ky$ is one-dimensional abelian. The quotient is spanned by $x+ky$, and [L2] gives $[x+ky,x+ky]=0+ky$, so it too is one-dimensional abelian. Hence both kernel and quotient are nilpotent by [L3].

Steps 1.1, 1.2, and 2.1 satisfy the hypotheses of a nilpotent-by-nilpotent extension and explicitly fail its proposed nilpotence conclusion. The construction is finite and uses no choice principle.

The map $\rho$ is real-linear. Since $\mathfrak{g}$ is abelian, $[at,bt]=0$, and since scalar multiples of $J$ commute, $[aJ,bJ]=0$. Hence $\rho ([at,bt])=[\rho (at),\rho (bt)]$ for all $a,b\in \mathbb{R}$, so [L1] makes $V$ a representation.

Suppose $W$ were a nonzero proper $\mathfrak{g}$-stable subspace of $V$. As $V$ has dimension two, $W=\mathbb{R}v$ for some $v\mathrm{\ne }0$. Stability gives $Jv=\lambda v$ for a real scalar $\lambda$. But $J^2=-I$, so $-v=J^{2}v={\lambda }^{2}v$, forcing ${\lambda }^2=-1$, impossible over $\mathbb{R}$. Thus no such $W$ exists, and [L2] shows that $V$ is irreducible.

The acting Lie algebra is abelian and the verified irreducible module has real dimension two, so it is the required counterexample to any one-dimensional conclusion over $\mathbb{R}$. Over $\mathbb{C}$, the same matrix acquires eigenlines for the eigenvalues $i$ and $-i$, pinpointing the missing field hypothesis. The finite calculation uses no choice principle.

The only nonzero basis bracket of $\mathfrak{a}$ spans $ky$, and $[ky,ky]=0$. Hence ${\mathfrak{a}}^{(1)}=ky$ and ${\mathfrak{a}}^{(2)}=0$, so $\mathfrak{a}$ is solvable by [L2].

For $0\le i<p-1$, $(XY-YX)v_i=(i+1-i)v_{i+1}=Y{v}_i$. At the wraparound index, $(XY-YX)v_{p-1}=(0-(p-1))v_0=v_0=Y{v}_{p-1}$ in characteristic $p$; this includes $p=2$. Thus $[X,Y]=Y$, and the other basis-bracket identities are automatic, so $x\mapsto X$, $y\mapsto Y$ is a representation by [L3].

Let $0\mathrm{\ne }W\subseteq V$ be stable under $X$ and $Y$, and choose a nonzero $w={\sum }_j{c}_j{v}_j\in W$ with $c_i\mathrm{\ne }0$. The scalars $0,1,\dots ,p-1$ are distinct in $k$. Therefore the Lagrange polynomial $P_i(T)={\prod }_{j\mathrm{\ne }i}(T-j)\mathrm{/}(i-j)$ is defined and satisfies $P_i(X)w=c_i{v}_i\in W$. Hence $v_i\in W$. Repeated application of the cyclic operator $Y$ puts every basis vector $v_j$ in $W$, so $W=V$. By [L4], $V$ is irreducible.

Since $p\ge 2$, this irreducible module has dimension greater than one. A common eigenvector would span a nonzero proper stable line, contradicting step 2.1. Thus the solvable algebra in step 1.1 and the representation in step 1.2 violate the common-eigenvector conclusion when characteristic zero is removed from [L1]. Every selection is from fixed finite coordinates, so no choice principle is used.

Set $V_i=\mathrm{span}(e_1,\dots ,e_i)$ for $0\le i\le n$. If $A$ is strictly upper triangular, its $j$th column has nonzero entries only in rows smaller than $j$, so $A{e}_j\in V_{j-1}$. Therefore $A(V_i)\subseteq V_{i-1}$ for every $i$. Iterating gives $A^n(V)=A^n(V_n)\subseteq V_0=0$, so every $A\in {\mathfrak{n}}_n(k)$ is a nilpotent operator.

Taking $i=1$ in step 1.1 gives $A{e}_1=0$ for every $A\in {\mathfrak{n}}_n(k)$, so $e_1\mathrm{\ne }0$ is a common zero vector. The full chain $0=V_0\subset V_1\subset \cdots \subset V_n=V$ has $\dim V_i=i$ and the containment required by [L1], so this explicit standard flag realizes Engel triangularization.

If $n=1$, then ${\mathfrak{n}}_1(k)=0$, the sole operator is zero, and the flag $0\subset k{e}_1$ satisfies the same calculation. For $n\ge 2$, [L2] additionally identifies the acting Lie algebra as nilpotent of class $n-1$, although elementwise nilpotence of this particular representation was proved directly in step 1.1. All basis and flag data are explicit and finite, so no choice principle is used.